LeetCode 之 Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

1
/ \
2   2
/ \ / \
3  4 4  3

But the following is not:

1
/ \
2   2
\   \
3    3

Note:

Bonus points if you could solve it both recursively and iteratively.

详细的代码如下：

1）递归解决：

bool isSymmetric(TreeNode* root) {
if(root==NULL)
return 1;
return isSymmetric(root->left,root->right);
}
bool isSymmetric(TreeNode* left,TreeNode* right){
if(!left&&!right)
return 1;
if(!left&&right||left&&!right||left->val!=right->val)
return 0;
return isSymmetric(left->left,right->right)&&isSymmetric(left->right,right->left);
}

分析：对于左右节点，如果都为空返回1，都不为空且val相同，递归比较左-左 ：右-右和左-右 ：右-左，这两对子树。

2）迭代解决，要选择一种合适的数据结构，我用的是队列：

bool isSymmetric(TreeNode* root){
if(!root)
return 1;
TreeNode* leftNode=root->left;
TreeNode* rightNode=root->right;
if(!leftNode&&!rightNode)
return 1;
if(!leftNode&&rightNode||leftNode&&!rightNode||leftNode->val!=rightNode->val)
return 0;
queue<TreeNode*> left,right;
left.push(leftNode);right.push(rightNode);
while(!left.empty()&&!right.empty()){
if(left.front()->val==right.front()->val){
leftNode=left.front();rightNode=right.front();
left.pop();right.pop();
if(leftNode->left==NULL&&rightNode->right!=NULL||leftNode->left!=NULL&&rightNode->right==NULL){
return 0;
}
else if(leftNode->left!=NULL && rightNode->right!=NULL){
left.push(leftNode->left);right.push(rightNode->right);
if(leftNode->right!=NULL&&rightNode->left!=NULL){
left.push(leftNode->right);right.push(rightNode->left);
}else if(leftNode->right!=NULL&&rightNode->left==NULL||leftNode->right==NULL&&rightNode->left!=NULL){
return 0;
}
}
else{
if(leftNode->right!=NULL&&rightNode->left!=NULL){
left.push(leftNode->right);right.push(rightNode->left);
}else if(leftNode->right!=NULL&&rightNode->left==NULL||leftNode->right==NULL&&rightNode->left!=NULL){
return 0;
}
}
}
else
return 0;
}
return left.empty()==right.empty();
}

有些地方比较繁琐，还可以优化。大致的意思是：左右两个子树分别用left和right队列存起来，循环对比两个队列的头元素leftNode和rightNode如果val相同：则以pop出这两个节点，同时以镜像的方式把leftNode的左右子节点和rightNode的左右子节点入队；如果不同：返回false。最后在循环判断如果left和right同时为空返回true否则返回false。